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- a. Draw a three-dimensional structure for the following steroid. b. What is the structure of the single stereoisomer formed by reduction of this ketone wrth H2, Pd-C? Explain why only one stereoisomer is formed.Draw the products formed (including stereoisomers) when attached compound is reduced with NaBH4 in CH3OH.Identify A–E.
- Thank you for the answer, would it be possible to follow the path I have attached as well srarting from methanol ---> CH3Br -----> CH3MgBr-------> (H3C)CH-CH2OH ----> E1 IsobutaleneWhich of the following pairs are keto–enol tautomers? a. CH3CH2CH=CHCH2OH and CH3CH2CH2CH2CH=O b. CH3CH-OHCH3 and CH3C=OCH3 c. CH3CH2CH=CHOH and CH3CH2CH2CH=O d. CH3CH2CH2CH=CHOH and CH3CH2CH2C=OCH3 e. CH3CH2CH2HO-C=CH2 and CH3CH2CH2C=OCH3Give two syntheses for (CH3)2CH¬O¬CH2CH3, and explain which synthesis is better
- Explain the following result. Acetic acid (CH3COOH), labeled at its OH oxygen with theuncommon 18O isotope (shown in red), was treated with aqueous base, and then the solution was acidified. Two products having the 18O label at different locations were formed.Oxidation of cholesterol converts the alcohol to a ketone. Under acidic or basic oxidationconditions, the C“C double bond migrates to the more stable, conjugated position. BeforeIR and NMR spectroscopy, chemists watched the UV spectrum of the reaction mixture tofollow the oxidation. Describe how the UV spectrum of the conjugated product, cholest-4-en-3-one, differs from that of cholesterolExplain the following result. Acetic acid (CH3COOH), labeled at its OH oxygen with theuncommon 18O isotope (shown in red), was treated with aqueous base, and then the solution was acidied. Two products having the 18O label at different locations were formed.